给定数字N ,任务是找到具有N个数字的最小三角数的索引。
A number is termed as a triangular number if we can represent it in the form of a triangular grid of points such that the points form an equilateral triangle and each row contains as many points as the row number, i.e., the first row has one point, the second row has two points, the third row has three points and so on. The starting triangular numbers are 1, 3, 6, 10, 15, 21, 28…………
例子:
Input: N = 2
Output: 4
Smallest triangular number with 2 digits = 10, and 4 is the index of 10.
Input: N = 3
Output: 14
Smallest triangular number with
3 digits = 105, and 14 is the index of 105.
方法:对该问题的主要观察是,具有N位数字的最小三角数的索引形成一个序列,该序列为–
1, 4, 14, 45, 141...
这带有N位数字的最小三角数的索引项将是
下面是上述方法的实现:
C++
// C++ implementation of
// the above approach
#include
using namespace std;
// Function to return index of smallest
// triangular no n digits
int findIndex(int n)
{
float x = sqrt(2 * pow(10, (n - 1)));
return round(x);
}
// Driver Code
int main()
{
int n = 3;
cout << findIndex(n);
return 0;
}
Java
// Java implementation of the above approach
class GFG{
// Function to return index of smallest
// triangular no n digits
static double findIndex(int n)
{
double x = Math.sqrt(2 * Math.pow(10, (n - 1)));
return Math.round(x);
}
// Driver code
public static void main(String[] args)
{
int n = 3;
System.out.print(findIndex(n));
}
}
// This code is contributed by shubham
Python3
# Python3 implementation of
# the above approach
import math
# Function to return index of smallest
# triangular no n digits
def findIndex(n):
x = math.sqrt(2 * math.pow(10, (n - 1)));
return round(x);
# Driver Code
n = 3;
print(findIndex(n));
# This code is contributed by Code_Mech
C#
// C# implementation of the above approach
using System;
class GFG{
// Function to return index of smallest
// triangular no n digits
static double findIndex(int n)
{
double x = Math.Sqrt(2 * Math.Pow(10, (n - 1)));
return Math.Round(x);
}
// Driver code
public static void Main(String[] args)
{
int n = 3;
Console.Write(findIndex(n));
}
}
// This code is contributed by AbhiThakur
Javascript
14